NCERT Solutions for Exercise 3.2, Class 11, Maths

Exercise 3.2 Trigonometric Functions
Question and Answers
Class 11 – Maths

Class	Class 11
Subject	Mathematics
Chapter Name	Trigonometric Functions
Chapter No.	Chapter 3
Exercise	Exercise 3.2
Category	Class 11 Maths NCERT Solutions

Question 1 Find the values of other five trigonometric functions if

(1) cos x = – 1 / 2 , x lies in third quadrant.

Answer

(1) Cos x = -1/2

sec x = -2
sin²x+cos²x=1
sin²x=1-cos²x
Substituting  cos x = – 1/ 2 in the formula, we obtain,
sin²x = 1 – (-1/2)²
sin²x = 1 – (1/4)
sin²x = 3/4
Sin x = ±√ 3 / 2
Since x lies in the 3rd quadrant, the value of sin x will be negative.
Sin x = -√ 3 / 2
Cosec x = 1 / sin x
Cosec x = 1 / (- √ 3 / 2)
Cosec x = – 2 / √ 3
tan x = sin x / cos x
tan x = ( -√ 3 / 2 ) / (-1/2)
tan x = √3
cot x = 1/ tan x
cot x = 1 / √3

(2) sin x = 3/ 5 , x lies in second quadrant

Sin x = 3/5
cosec x = 1/ sin x
cosec x = 1 /(3/5)
cosec x = 5/3
sin²x+cos²x=1
cos²x=1- sin²x
Substituting  sin x = 3 / 5 in the formula, we obtain,
cos²x=1 – (3/5)²
cos²x=1 – (9/25)
cos²x=16 /25
cos x = ± 4 / 5
Since x lies in the second quadrant, the value of cos x will be negative.
cos x = – 4 / 5
sec x = 1 / cos x
sec x =  1/ (-4/5)
sec x = – 5/4
tan x = sin x / cos x
tan x = (3/5) / (-4/5)
tan x = -3/ 4
cot x = 1 / tan x
cot x = 1 / (-3/4)
cot x = -4 /3

(3) cot x = 4/3 , x lies in third quadrant

cot x = 3/ 4
tan x = 1 / cot x
tan x = 4/3
sec²x – tan²x = 1
sec²x = 1+ tan²x
sec²x = 1+ (4/3)²
sec²x = 1+ (16 /9)
sec²x = 25 / 9
sec x = ± 5 / 3
Since x lies in the 3rd quadrant, the value of sec x will be
sec = – 5/3
cos x = 1 / sec x
cos x = 1 / (-5/3)
cos x = – 3/ 5
tan x = sin x/ cos x
sin x = tan x × cos x
sin x = ( 4/ 3) × ( – 3/ 5 )
sin x = – 4/ 5
cosec x = 1/ sin x
cosec x = -5/4

(4) sec x = 13/5 , x lies in fourth quadrant

sec x = 13 / 5
cos x = 1 /sec x
cos x = 1 /(13 /5 )
cos x = 5/ 13
sec²x – tan²x = 1
tan²x = sec²x – 1
tan²x = (13 / 5)² – 1
tan²x = (169 / 25 ) -1
tan²x =  144 / 25
tan x = ± 12 / 5
Since x lies in the 4th quadrant the value of tan x will be negative.
tan x = -12/ 5
cot x = 1 / tan x
cot x = – 5 /12
tan x = sin x /cos x
sin x = tan x × cos x
sin x  = (-12/ 5) × ( 5/13)
sin x= -12 /13
cosec x = 1 / sin x
cosec x =-  13 /12

(5) tan x = – 5 /12 , x lies in second quadrant.

tan x = – 5/12
cot x = 1 / tan x
cot x = 1 /(-5/12)
cot x = -12 /5
1 + tan² x = sec² x
We can write it as
1 + (-5/12)² = sec² x
Substituting the values
1 + 25/144 = sec² x
sec² x = 169/144
sec x = ± 13/12
Here x lies in the second quadrant so the value of sec x will be negative
sec x = – 13/12
cos x = 1 /sec x
cos x = – 12 /13
tan x = sin x/ cos x
sin x = tan x × cos x
sin x = (-5/12) × (-12 / 13)
sin x= 5 /13
cosec x= 1 / sin x
cosec x = 13 / 5

Find the values of the trigonometric functions in Exercises

(6) sin 765°

(6) We know that the values of sin x repeat after an interval of 2π  or 360°
sin 765° = sin (2 × 360°+45°)
sin 765° = sin 45°
sin 765° = 1/ √2

(7) cosec (– 1410°)

(7) We know that the values of sin x repeat after an interval of 2π  or 360°
cosec (-1410° ) = cosec (-1410°+4 × 360°)
= cosec(-1410°+1440°)
= cosec30°
= 2

(8) tan 19π /3

(8) We know that the values of sin x repeat after an interval of 2π  or 360°

(9) sin (– 11π/ 3 )

(9) We know that the values of sin x repeat after an interval of 2π  or 360°

(10) cot (– 15π/ 4 )

(10) We know that the values of sin x repeat after an interval of 2π  or 360°